Recall the normal ordering of bosonic operators in QFT is defined by a re-arrangement of operators to put creation operators to the left of annihilation operators in the product. This is designed to avoid accidentally annihilating $|0\rangle$ when looking at an expectation value in relation to the vacuum state.
$$
: \hat{b}^\dagger\hat{b} : \: =\: \hat{b}^\dagger\hat{b} \\ : \hat{b}\hat{b}^\dagger: \: = \: \hat{b}^\dagger\hat{b}
$$
In CFTs, I've seen defined the normal ordering of operators as the zeroth basis field of the Laurent expansion of the radial ordering product.
$$\mathcal{R}(a(z)b(w)) = \sum_{n = -n_0}^\infty (z-w)^n P_n(w),$$
and select
$$P_0(w) = \: : a(w)b(w) : $$
Is there an equivalence between these two definitions? What is the CFT analog of not annihilating the vaccuum/ how do we show this definition has that property?
Best Answer
In Quantum Field theory, for non-interacting fields, the normal ordering can be defined by requiring that the product of the two fields doesn't have the singular part. Since for non-interacting fields the singular part is nothing but the Vacuum expactation value (and is just 1 term), it is sufficient to write: $$:\phi^2:\,\,\,=\phi^2-\langle\phi\phi\rangle$$
In CFT we can't just do that. Take the energy momentum tensor. It's OPE is known to be: $$T(z)T(w)=\frac{c/2}{(z-w)^4}+\frac{2T(w)}{(z-w)^2}+\frac{\partial T(w)}{(z-w)}+regular\,terms$$ if we try to take out $\langle T(z)T(w)\rangle$, we obtain: $$T(z)T(w)-\langle T(z)T(w)\rangle=\frac{2T(w)}{(z-w)^2}+\frac{\partial T(w)}{(z-w)}+regular\,terms$$ which is still singular.
Then, instead of subtracting only the V.A.V. , every non singular term is taken out. If we have two operators with the following OPE: $$A(z)B(w)=\sum_{n-\infty}^N \frac{\{AB\}_n(w)}{(z-w)^n}$$ with $N$ positive integers (which means that the number of singular parts can be finite) and $\{AB\}_n(w)$ the resulting fields of the expansion. We then define the normal ordered product as: $$(AB)(w):=\{AB\}_0(w)$$ In fact, we can define the Contraction as: $$C(A(z)B(w)):=\sum_{n=1}^{N}\frac{\{AB\}_n(w)}{(z-w)^n}$$ And then the normal ordered product is just: $$(AB)(w)=\lim_{z\to w}[A(z)B(w)-C(A(z)B(w))]$$ since all the terms $\{AB\}_n(z-w)^n$ with $n>0$ goes to zero as $z\to w$.
In this context, we can give an integral representation of this normal ordered product as: $$(AB)(z)=\oint_z\frac{dw}{2\pi i}\frac{A(w)B(z)}{w-z}$$ where the contour integral contains the point $z$.